An immersion heater rated 2.0A, 240V is used to boil water from temperature 52°c to 100°c. If the mass of the water is 2.5kg. determine the time taken to bo...
An immersion heater rated 2.0A, 240V is used to boil water from temperature 52°c to 100°c. If the mass of the water is 2.5kg. determine the time taken to boil the water. [specific heat capacity of water = 4.2 x 103 jkg-1 k-1]
Answer Details
The equation for the energy required to heat a substance can be written as:
Q = mcΔT
where Q is the energy required, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
Using the given information, we can solve for the energy required to heat the water from 52°C to 100°C:
ΔT = 100°C - 52°C = 48°C
Q = (2.5 kg)(4.2 x 10^3 J/kg°C)(48°C) = 5.04 x 10^5 J
The power of the immersion heater is given as 2.0A at 240V, so the power output can be calculated as:
P = IV = (2.0A)(240V) = 480W
The time required to provide the energy needed to heat the water can be found using the formula:
t = Q/P
t = (5.04 x 10^5 J) / (480W) = 1.05 x 10^3 s
Therefore, the time taken to boil the water is 1.05 x 10^3 seconds, which is equivalent to 17.5 minutes (since there are 60 seconds in a minute).
The answer closest to this value is "1.05 x 10^3 s".